What is the slope of the line tangent to $f(x) = x^{2}+3x-3$ at $x = 1$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{((x+\Delta x)^{2}+3(x+\Delta x)-3) - (x^{2}+3x-3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(x^{2}+2x \Delta x+\Delta x^{2}+3(x+\Delta x)-3) - (x^{2}+3x-3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{x^{2}+2(x \Delta x)+\Delta x^{2}+3x+3(\Delta x)-3-x^{2}-3x+3}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2(x \Delta x)+\Delta x^{2}+3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} 2x+\Delta x+3$ $ = 2x+3$ $ = (2)(1)+3$ $ = 5$